(1)
Photo-electrons with minimum velocity
imply that K min applies here so indicates the wavelength of incident light
(l = 0.50 mm) is at the cutoff wavelength so: l = l c Then the cutoff or threshold frequency is:
f
c = c /l
c = (3 x 108 ms-1)/
(500 x 10-9 m) = 6 x 10 14 s-1
Then
work function would be: f = hf c =
(6.62 x 10 -34 Js) ( 6 x 10 14 s-1)
(6.62 x 10 -34 Js) ( 6 x 10 14 s-1)
= 3.97 x 10 -19
J =
(3.97 x 10 -19 J )/ 1.60 x 10 -19
J /eV) = 2.48 eV
(2)
Stopping potential Vs = 0.54 V and l = 750 nm
f= c/ l =
(3 x 108 ms-1)/ (750 x 10-9 m) = 4 x 10 14 s-1
The
work function is obtained from: eVs = hf
- f
So,
transposing: f = hf
- eVs =
(6.62 x 10 -34 Js) (4 x 10 14 /s) - (1.6
x 10 -19 C) 0.54 J/C]
f = 2.64 x 10 -19 J -
0.86 x 10 -19 J =
1.78 x 10 -19 J
In
eV: f
= 1.78 x 10 -19
J / (1.6 x 10 -19 J/
eV) = 1.11 eV
(3)
K max = 0.57 eV
and photo-electrons dislodged from a metal surface by incident radiation
with l = 500 nm.
The
incident energy E = hf = h c/l =
(6.62 x 10 -34 Js) (3 x 108 ms-1)/ (500 x 10-9 m)
(6.62 x 10 -34 Js) (3 x 108 ms-1)/ (500 x 10-9 m)
h c/ l = 3.97 x 10-19 J
K
max = 0.57 eV (1.6 x 10 -19 J/ eV) = 0.91 x 10 -19
J
Therefore:
K
max = 0.91 x
10 -19 J
= [3.97 x 10-19 J - f ]
So,
the work function is: f =
[3.97 x 10-19 J -
0.91 x 10 -19
J] =
3.06 x 10 -19
J
To
get in eV: (3.06 x 10 -19 J)/ (1.6 x 10 -19 J/ eV) = 1.91 eV
The
stopping potential in volts:
Vs
= hf/ e
- f/e where the slope h/e = 4.13 x 10-15
Js/C
The
frequency f = c/l = (3
x 108 ms-1)/ (500 x
10-9 m) =
6 x 10 14 c/s
6 x 10 14 c/s
Then:
Vs
= (4.13 x 10-15 J s/C )(6 x 10 14 s-1)
-
(3.06 x 10 -19 J)/ (1.6 x 10 -19 C)
(3.06 x 10 -19 J)/ (1.6 x 10 -19 C)
Vs = 2.47 V – 1.91 V = 0.56 V
No comments:
Post a Comment